A number theory problem by Dev Sharma

$18^{30}=-1=24(mod5)$ $18^{30}=0=24(mod8)$

so, $18^{30}=24(mod40)$ . Since we know that $37^{\phi(100)}=1(mod100)$ , therefore

$37^{18^{30}}=37^{24}=37^{8+16}=61(mod100)$ .

The reason I break 24 into 8+16, so that we can reduce the work of computation, just find $37^1,37^2,37^4,37^8,37^{16}$ and then only multiply the last two numbers.

Note, in fact, $37^{20}=1(mod100)$ but for this solution, we don't need to find the lowest power.

All options in the multiple choice have distinct last algarisms from each other. So, we can find the last algarism of $37^{18^{30}}$ :

Let's call LA(x) the function wich gives the last algarism of x.

So,

$LA(37^{18})=LA(37^{4\times4 + 2} = LA(37^{2})=9$

And,

$LA(37^{18^{30}}) = LA(9^{30}) = LA(9^{15\times2+0})=LA(9^{0})=1$

So the only possible choice is 61.